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Re: [EMHL] From my geometry textbook (and School folklore)

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  • xpolakis@otenet.gr
    Dear Nikos ... Yes, it s the same. ... That is, the circumcircle of ABC. C A P B B C Let P = BB / CC trABC ~ trAB C == angBAC =
    Message 1 of 2 , Jul 14, 2001
      Dear Nikos

      >>Let ABC be a triangle and consider an arbitrary triangle AB'C' such that:
      >>AB'C' ~ [=similar to] ABC [the vertex A is in common].
      >>Find the locus of the intersection of the lines BB' and CC'.
      >>(I. Ioannides: Mathematics of D' of Gymnasium. Vol II.
      >>OEDB: Athens, 1968, p. 52, #66)

      >At first glance, if the order of AB, AC is the
      >same as the order of AB', AC',

      Yes, it's the same.

      >we can say that if s = AC/AB then
      >a rotation around A by angle BAC moves B to B''
      >and the homothetic of B'' with ratio s is the point C.
      >The same synthesis of tranformations moves B' to C'
      >and hence BB' and CC' form an angle equal to angle BAC.
      >So the locus of the intersection of BB' and CC' is a
      >circular arc (and the symmetric about BC) every point
      >of which sees BC with a constant angle BAC.

      That is, the circumcircle of ABC.


      A P

      B C

      Let P = BB' /\ CC'

      trABC ~ trAB'C' ==>

      angBAC = angB'AC' ==> angBAB' = angCAC' (1)


      AB / AC = AB' / AC' or AB / AB' = AC / AC' (2)

      (1) and (2) ==> trABB' ~ trACC' ==> angAB'B = angAC'C ==>
      quadAB'PC' is cyclic ==> angBPC = angB'AC' = angBAC ==>
      The locus of P is the circumcircle of ABC.


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